Scala applying a PartialFunction with () is not the same as .apply()
我想在一个项目(Play Framework 2.4)中重构我的 scala 代码时，我想到了这个想法：
(为了提供一个最小的工作示例，我已经更改了一些类，例如，我分别用 Int 和 Option[Int] 更改了 Result 和 Future[Result])
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object ParFuncApply {
trait CanBeAuthenticatedRequest[A]
trait AuthenticatedRequest[A] extends CanBeAuthenticatedRequest[A]
trait UnauthenticatedRequest[A] extends CanBeAuthenticatedRequest[A]
private def fold[T](authenticated: (AuthenticatedRequest[_]) => T)
(unauthenticated: (UnauthenticatedRequest[_]) => T):
PartialFunction[CanBeAuthenticatedRequest[_], T] = {
case ar: AuthenticatedRequest[_] => authenticated(ar)
case ur: UnauthenticatedRequest[_] => unauthenticated(ur)
}
def apply(request: CanBeAuthenticatedRequest[_])
(authenticated: (AuthenticatedRequest[_]) => Int)
(unauthenticated: (UnauthenticatedRequest[_]) => Int): Int = {
fold(authenticated)(unauthenticated)(request)
}
def async(request: CanBeAuthenticatedRequest[_])
(authenticated: (AuthenticatedRequest[_]) => Option[Int])
(unauthenticated: (UnauthenticatedRequest[_]) => Option[Int]): Option[Int] = {
fold(authenticated)(unauthenticated)(request)
}
}
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上面的代码编译好了。
然后我：我应该将 fold[T] 参数化类型限制为 Int 和 Option[Int]，所以我添加了：
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object ParFuncApply {
trait CanBeAuthenticatedRequest[A]
trait AuthenticatedRequest[A] extends CanBeAuthenticatedRequest[A]
trait UnauthenticatedRequest[A] extends CanBeAuthenticatedRequest[A]
sealed trait Helper[T]
object Helper {
implicit object FutureResultHelper extends Helper[Option[Int]]
implicit object ResultHelper extends Helper[Int]
}
private def fold[T: Helper](authenticated: (AuthenticatedRequest[_]) => T)
(unauthenticated: (UnauthenticatedRequest[_]) => T):
PartialFunction[CanBeAuthenticatedRequest[_], T] = {
case ar: AuthenticatedRequest[_] => authenticated(ar)
case ur: UnauthenticatedRequest[_] => unauthenticated(ur)
}
def apply(request: CanBeAuthenticatedRequest[_])
(authenticated: (AuthenticatedRequest[_]) => Int)
(unauthenticated: (UnauthenticatedRequest[_]) => Int): Int = {
fold(authenticated)(unauthenticated)(request)
}
def async(request: CanBeAuthenticatedRequest[_])
(authenticated: (AuthenticatedRequest[_]) => Option[Int])
(unauthenticated: (UnauthenticatedRequest[_]) => Option[Int]): Option[Int] = {
fold(authenticated)(unauthenticated)(request)
}
}
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但是，如果我更改，则此代码不再编译：
fold(authenticated)(unauthenticated)(request) 到 fold(authenticated)(unauthenticated).apply(request)(我添加了对 apply() 的显式调用)它编译。为什么会这样？在一个类上调用 () 和 .apply() 应该是一样的，不是吗？
编译器似乎要求将返回类型(Int 或 Option[Int])传递给 PartialFunction，而不是 CanBeAuthenticatedRequest 类型。
因为您在 `fold[T : Helper]\\’ 中定义了一个上下文绑定，编译器将添加另一个参数列表。换句话说，上下文绑定只是语法糖：
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private def fold [T ](authenticated : (AuthenticatedRequest [_]) => T )
(unauthenticated : (UnauthenticatedRequest [_]) => T )
(implicit helper : Helper [T ): PartialFunction [CanBeAuthenticatedRequest [_], T ]
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所以当你调用
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fold(authenticated)(unauthenticated)(request)
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编译器认为 request 应该是显式指定的隐式 Helper[T].
- @vicaba 另外，另一个常见的解决方法是在调用之前先将部分函数分配给 val，而不是调用 apply。例如。 val f = fold(authenticated)(unauthenticated) f(request)
